CF #52 C Circular RMQ (segment tree Interval Update)

CF #52 C Circular RMQ (segment tree Interval Update)

Description

You are given circular arrayA_{0 ,?A1 ,?...,?AN? -? 1. There are two types of operations with it:}

Inc( Lf,? Rg,? V)-This operation increases each element on the segment [ Lf,? Rg] (Inclusively) V; Rmq( Lf,? Rg)-This operation returns minimal value on the segment [ Lf,? Rg] (Inclusively ).

Assume segments to be circular, so ifN? =? 5 andLf? =? 3 ,?Rg? =? 1, it means the index sequence: 3 ,? 4 ,? 0 ,? 1.

Write program to process given sequence of operations.

Input

The first line contains integerN(1? ≤?N? ≤? (200000). The next line contains initial state of the array:A_{0 ,?A1 ,?...,?AN? -? 1 (? -? 10^{6? ≤?AI? ≤? 106 ),AIAre integer. The third line contains integerM(0? ≤?M? ≤? 200000 ),M-The number of operartons. NextMLines contain one operation each. If line contains two integerLf,?Rg(0? ≤?Lf,?Rg? ≤?N? -? 1) it meansRmqOperation, it contains three integersLf,?Rg,?V(0? ≤?Lf,?Rg? ≤?N? -? 1 ;? -? 106? ≤?V? ≤? 106 )-IncOperation.}}

Output

For eachRmqOperation write result for it. please, do not use % lld specificator to read or write 64-bit integers in C ++. it is preffered to usecout (also you may use % I64d ).

Sample Input

Input

41 2 3 443 03 0 -10 12 1

Output

100

The question is easy to understand.

If a> B, query 0 ~ B and ~ N-1. The rest is the line segment tree Interval Update template, which determines whether c exists in m queries. For details, refer to the Code for better understanding.

#include 
 
  #include 
  
   #include #include 
   
    #include 
    
     #include 
     
      #define lson o << 1, l, m#define rson o << 1|1, m+1, rusing namespace std;typedef __int64 LL;const __int64 MAX =  9223372036854775807;const int maxn = 200010;int n, a, q, c, b;char str[1200];LL mi[maxn<<2], add[maxn<<2];void up(int o) {    mi[o] = min(mi[o<<1], mi[o<<1|1]);}void down(int o) {    if(add[o]) {        add[o<<1] += add[o];        add[o<<1|1] += add[o];        mi[o<<1] += add[o];        mi[o<<1|1] += add[o];        add[o] = 0;    }}void build(int o, int l, int r) {    if(l == r) {        scanf("%I64d", &mi[o]);        return;    }    int m = (l+r) >> 1;    build(lson);    build(rson);    up(o);}void update(int o, int l, int r) {    if(a <= l && r <= b) {        add[o] += c;        mi[o] += c;        return ;    }    down(o);    int m = (l+r) >> 1;    if(a <= m) update(lson);    if(m < b ) update(rson);    up(o);}LL query(int o, int l, int r) {    if(a <= l && r <= b) return mi[o];    down(o);    int m = (l+r) >> 1;    LL res = MAX;    if(a <= m) res = query(lson);    if(m < b ) res = min(res, query(rson));    return res;}int main(){    scanf("%d", &n);    build(1, 0, n-1);    scanf("%d", &q); getchar(); while(q--) {        gets(str);        if(sscanf(str,"%d %d %d", &a, &b, &c) == 3)  {            if(a <= b) update(1, 0, n-1);            else {                int tmp = b;                b = n-1; update(1, 0, n-1);                a = 0, b = tmp; update(1, 0, n-1);            }        } else {            LL ans ;            if(a <= b) ans = query(1, 0, n-1);            else {                int tmp = b;                b = n-1; ans = query(1, 0, n-1);                a = 0, b = tmp; ans = min(ans, query(1, 0, n-1));            }            printf("%I64d\n", ans);        }    }    return 0;}
     
    
   
  
 

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